Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Mock AIME 1 2006-2007 Problems/Problem 2"

(→‎Solution)
m
 
(2 intermediate revisions by 2 users not shown)
Line 5: Line 5:
 
----
 
----
  
−
*[[Mock AIME 1 2006-2007/Problem 1 | Previous Problem]]
+
*[[Mock AIME 1 2006-2007 Problems/Problem 1 | Previous Problem]]
  
−
*[[Mock AIME 1 2006-2007/Problem 3 | Next Problem]]
+
*[[Mock AIME 1 2006-2007 Problems/Problem 3 | Next Problem]]
  
 
*[[Mock AIME 1 2006-2007]]
 
*[[Mock AIME 1 2006-2007]]
 +
 +
[[Category:Intermediate Combinatorics Problems]]

Latest revision as of 14:53, 3 April 2012

Let $\star (x)$ be the sum of the digits of a positive integer $x$. $\mathcal{S}$ is the set of positive integers such that for all elements $n$ in $\mathcal{S}$, we have that $\star (n)=12$ and $0\le n< 10^{7}$. If $m$ is the number of elements in $\mathcal{S}$, compute $\star(m)$.

Solution

Equivalently, we need to place 12 indistinguishable balls into 7 distinguishable boxes so that no box contains more than 9 balls. There are ${12 + 7 - 1 \choose 7 - 1} = {18 \choose 6} = 18,564$ ways to place 12 objects into 7 boxes. Of these, 7 place all 12 into a single box. $7 \cdot 6 = 42$ place 11 in one box and 1 in a second. $7 \cdot 6 = 42$ place 10 into one box and 2 into a second. $7 \cdot \frac{6\cdot 5}{2} = 105$ place 10 into one box and 1 into each of two others. Thus, this gives us $m = 18564 - 7 - 42 - 42 - 105 = 18368$ so $\star(m) = 1 + 8 + 3 + 6 + 8 = 026$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_1_2006-2007_Problems/Problem_2&oldid=46000"